ZOHO Off Campus Coding Questions And Answers 2024, Check ZOHO Off Campus Coding Interview Questions

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ZOHO Off Campus Coding Questions And Answers 2024: All of the fundamental algorithms, data structures, and programming concepts are included in the ZOHO Off Campus Coding Interview Questions. We can assist you if you’re interested in working at ZOHO but are unsure about how to get ready for the questions in the ZOHO Off Campus Coding Interview Round. One of the most important sections of the Zoho Off Campus Interview exam is the coding round. In this article, we covered a variety of sample ZOHO Off campus Coding Questions and Answers 2024. We will talk about ZOHO Off Campus Interview Coding Questions 2024/Off Campus Coding Questions ZOHO in the below section.

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ZOHO Off Campus Drive Coding Interview Questions and Answers 

Question 1: Alex works at a clothing store. There is a large pile of socks that must be paired by color for sale. Given an array of integers representing the color of each sock, determine how many pairs of socks with matching colors there are.

For example, there are n=7 socks with colors ar = {1,2,1,2,1,3,2}. There is one pair of color 1 and one of color 2. There are three odd socks left, one of each color. The number of pairs is 2.

Function Description
Complete the sockMerchant function in the editor below. It must return an integer representing the number of matching pairs of socks that are available.
sockMerchant has the following parameter(s):
             n: the number of socks in the pile
             ar: the colors of each sock

Input Format
The first line contains an integer n, the number of socks represented in ar.
            The second line contains n space-separated integers describing the colors ar[i] of the socks in the pile.

Constraints
1 <= n <= 100
1 <= ar[i] <= 100 & 0 <= i < n

Output Format
Return the total number of matching pairs of socks that Alex can sell.

Sample Input
9
10 20 20 10 10 30 50 10 20
Sample Output
             3

Explanation
Alex can match 3 pairs of socks i.e 10-10, 10-10, 20-20
while the left out socks are 50, 60, 20

C PROGRAM 

#include<stdio.h>
int sockMerchant(int n, int arr[])
{
int freq[101]={0};
int ans = 0,i;
for(i=0;i<n;i++)
freq[arr[i]]++;
for(i = 0; i <= 100; i++)
{
ans = ans+ freq[i]/2;
}
return ans;
}
int main ()
{
int n;
scanf("%d",&n);
int arr[101],i;
for (i = 0; i < n; i++)
{
scanf("%d",&arr[i]);
}

int ans=sockMerchant(n,arr);
printf("%d\n",ans);
return 0;
}

C++ PROGRAM 

#include<bits/stdc++.h>
using namespace std;
int sockMerchant(int n, int arr[])
{
    int freq[101]={0};
    int ans = 0;
    for(int i=0;i<n;i++)
    {
        int value=arr[i];
        freq[value]++;
    }   
    for(int i = 0; i <= 100; i++)
    {
        ans = ans+ freq[i]/2;
    }
return ans;
}
int main ()
{
int n;
cin >> n;
int arr[n]={0};
for (int i = 0; i < n; i++)
{
cin>>arr[i];
}
int res=sockMerchant(n,arr);
cout<<res<<endl;
return 0;
}

JAVA PROGRAM 

import java.util.*;
class Main
{
    public static int sockMerchant(int n, int arr[])
    {
         int freq[]=new int[101];
        for(int i=0;i<n;i++)
        {
            freq[arr[i]]++;
        }
        int ans=0;
        for(int i=0;i<=100;i++)
            ans=ans+freq[i]/2;
        return ans;
    }
    public static void main(String[] args)
    {
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        int arr[]=new int[n];
        for(int i=0;i<n;i++)
            arr[i]=sc.nextInt();
        int ans=sockMerchant(n,arr);
        System.out.println(ans);
}
}

PHYTHON PROGRAM 

def sockMerchant(n, ar):
    pairs = 0
    set_ar = set(ar)
    for i in set_ar:
        count = ar.count(i)
        pairs+=count//2
    return pairs
    
n = int(input())
ar = list(map(int, input().split()))
print(sockMerchant(n, ar))

Question 2: A left rotation operation on an array shifts each of the array’s elements unit to the left. For example, if 2 left rotations are performed on array [1, 2, 3, 4, 5], then the array would become [3, 4, 5, 1, 2].

Given an array of integers and a number, , perform left rotations on the array. Return the updated array to be printed as a single line of space-separated integers.

Function Description

Complete the function rotLeft in the editor below. It should return the resulting array of integers.

rotLeft has the following parameter(s):

  • An array of integers .
  • An integer , the number of rotations.

Input Format

The first line contains two space-separated integers and , the size of and the number of left rotations you must perform.

The second line contains space-separated integers a[i].

Constraints

  • 1 <= n <= 10^5
  • 1 <= d <= n
  • 1 <= a[i] <= 10^8

Output Format

Print a single line of space-separated integers denoting the final state of the array after performing d left rotations.

Sample Input
5 4
1 2 3 4 5

Sample Output
5 1 2 3 4

Explanation
When we perform d=4 left rotations, the array undergoes the following sequence of changes:

[1,2,3,4,5] → [2,3,4,5,1] → [3,4,5,1,2] → [4,5,1,2,3] → [5,1,2,3,4]

Test Case : 1

Input (stdin)

  • 5 4
  • 1 2 3 4 5

Expected Output

  • 5 1 2 3 4

Test Case : 2
Input (stdin)

  • 20 10
  • 41 73 89 7 10 1 59 58 84 77 77 97 58 1 86 58 26 10 86 51

Expected Output

  • 77 97 58 1 86 58 26 10 86 51 41 73 89 7 10 1 59 58 84 77

C PROGRAM 

#include<stdio.h> 
int rotLeft(int arr[], int n, int d)
{
    int i, j;
    int first;
    for(i=0; i<d; i++)
    {
        first = arr[0];
        for(j=0; j<n-1; j++)
        {
            arr[j] = arr[j+1];
        }
        arr[j] = first;
    }
    return *arr;
}
int main()
{
    int n, d, i;
    scanf("%d",&n);
    scanf("%d",&d);
    int list[n];
    for(i=0; i<n; i++)
    {
        scanf("%d",&list[i]);
    }
    rotLeft(list, n, d);
    for(i=0; i<n; i++)
    {
        printf("%d ",list[i]);
    }
}

C++ PROGRAM 

#include<bits/stdc++.h>
using namespace std;
int rotLeft(int arr[], int n, int d)
{
    int i, j;
    int first;
    for(i=0; i<d; i++)
    {
        first = arr[0];
        for(j=0; j<n-1; j++)
        {
            arr[j] = arr[j+1];
        }
        arr[j] = first;
    }
    return *arr;
}
int main()
{
    int n, d, i;
    cin>>n;
    cin>>d;
    int list[n];
    for(i=0; i<n; i++)
    {
        cin>>list[i];
    }
    rotLeft(list, n, d);
    for(i=0; i<n; i++)
    {
        cout<<list[i]<<" ";
    }
}

JAVA PROGRAM

import java.util.*;
class Main
{
    public static void rotateLeft(int a[],int n, int d)
    {
        int first,i,j;
        for(i=0;i<d;i++)
        {
            first=a[0];
            for(j=0;j<n-1;j++)
                a[j]=a[j+1];
            a[j]=first;
        }
    }
    public static void main(String[] args)
    {
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        int d=sc.nextInt();
        int a[]=new int[n];
        for(int i=0;i<n;i++)
            a[i]=sc.nextInt();
rotateLeft(a,n,d);

for(int i=0;i<n;i++)
System.out.print(a[i]+" ");
}
}

PHYTHON PROGRAM 

def rotateLeft(n,d,arr):
    for i in range(d):
        arr = rotatearr(n,arr)
    return arr
    
def rotatearr(n, arr):
    first = arr[0]
    for i in range(n-1):
        arr[i] = arr[i+1]
    arr[n-1] = first
    return arr

, d = map(int, input().split())
arr = list(map(int, input().split()))
for i in rotateLeft(n,d,arr):
    print(i, end=" ")

Question 3: 

A Pythagorean triplet is a set of three integers a, b and c such that a2 + b2 = c2. Given a limit, generate all Pythagorean Triples with values smaller than given limit.

Input : limit = 20
Output : 3 4 5
         8 6 10
         5 12 13
         15 8 17
         12 16 20

Simple Solution is to generate these triplets smaller than given limit using three nested loop. For every triplet, check if Pythagorean condition is true, if true, then print the triplet. Time complexity of this solution is O(limit3) where ‘limit’ is given limit.

An Efficient Solution can print all triplets in O(k) time where k is number of triplets printed. The idea is to use square sum relation of Pythagorean triplet, i.e., addition of squares of a and b is equal to square of c, we can write these number in terms of m and n such that,

       a = m2 - n2
       b = 2 * m * n
       c  = m2 + n2
because,
       a2 = m4 + n4 – 2 * m2 * n2
       b2 = 4 * m2 * n2
       c2 = m4 + n4 + 2* m2 * n2

We can see that a2 + b2 = c2, so instead of iterating for a, b and c we can iterate for m and n and can generate these triplets

C PROGRAM 

//  A C program to generate pythagorean triplets
// smaller than a given limit
#include 
#include 
//  Function to generate pythagorean triplets
//  smaller than limit
void pythagoreanTriplets (int limit)
{
    // triplet:  a^2 + b^2 = c^2
  int a, b, c = 0;
    //  loop from 2 to max_limitit
  int m = 2;
    // Limiting c would limit all a, b and c
  while (c < limit)
    {
    // now loop on j from 1 to i-1
      for (int n = 1; n < m; ++n)
    {
    // Evaluate and print triplets using
    // the relation between a, b and c
      a = m * m - n * n;
      b = 2 * m * n;
      c = m * m + n * n;
      if (c > limit)
        break;
      printf ("%d %d %d \n", a, b, c);
    }
      m++;
    }
}
// Driver program
int main ()
{
  int limit = 20;
  pythagoreanTriplets (limit);
  return 0;
}

C++ PROGRAM 

#include<bits/stdc++.h>
using namespace std;
//  Function to generate pythagorean triplets
//  smaller than limit
void pythagoreanTriplets (int limit)
{
    // triplet:  a^2 + b^2 = c^2
    int a, b, c = 0;
    //  loop from 2 to max_limitit
    int m = 2;
    // Limiting c would limit all a, b and c
while (c < limit)
{
// now loop on j from 1 to i-1
for (int n = 1; n < m; ++n)
{
// Evaluate and print triplets using
// the relation between a, b and c
a = m * m - n * n;
b = 2 * m * n;
c = m * m + n * n;
if (c > limit)
break;
cout<<a<<" "<<b<<" "<<c<<endl;
}
m++;
}
}
// Driver program
int main ()
{
int limit = 20;
pythagoreanTriplets (limit);
return 0;
}

JAVA PROGRAM 

import java.util.*;
class Main 
{
    public static void pythagoreanTriplets(int limit)
    {
        int a,b,c=0;
        int m=2;
        while(c<limit)
        {
            for(int n=1;n<m;++n)
            {
                a=m*m-n*n;
                b=2*m*n;
                c=m*m+n*n;
                if(c>limit)
                    break;
                System.out.println(a+" "+b+" "+c);
            }
            m++;
        }
    }
    public static void main(String[] args)
    {
        int limit=20;
        pythagoreanTriplets(limit);
    }
}

PHYTHON PROGRAM 

def pythagoreanTriplets(limit):
    a = b = c = 0
    m = 2
    while True:
        for n in range(1,m+1):
            a = m*- n*n
            b = 2*m*n
            c = m*+ n*n
            if c > limit:
                break
            if a==0 or b==0 or c==0:
                break
            print(a,b,c)
        m=m+1
        
limit = int(input())
pythagoreanTriplets(limit)

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